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aliina [53]
2 years ago
5

A distant astronomical object (a quasar) is moving away from us at half the speed of light. what is the speed of the light we re

ceive from this quasar?
Physics
1 answer:
Aloiza [94]2 years ago
7 0

For a distant astronomical object (a quasar) is moving away from us at half the speed of light, the speed of the light we receive from this quasar  is mathematically given as c = 3x108 m/s

<h3>What is the speed of light?</h3>

Generally, the equation for the   is mathematically given as

The speed of light can be said to be measured to be approximately the value c = 3x108 m/s

.

In conclusion, An will not influence  3x108 m/s unless both are inertia and frames refrences where newtons laws are valid, so the speed of light measured from the earth frame is equal to  3x108 m/s

Read more about Speed

brainly.com/question/4931057

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A team of astronauts is on a mission to land on and explore a large asteroid. In addition to collecting samples and performing e
GenaCL600 [577]

Answer:

463.4 m/s

Explanation:

The escape velocity on the surface of a planet/asteroid is given by

v=\sqrt{\frac{2GM}{R}} (1)

where

G is the gravitational constant

M is the mass of the planet/asteroid

R is the radius of the planet/asteroid

For the asteroid in this problem, we know

\rho=4.49\cdot 10^6 g/m^3 is the density

V=3.32\cdot 10^{12} m^3 is the volume

So we can find its mass:

M=\frac{\rho}{V}=(4.49\cdot 10^6 g/m^3)(3.32\cdot 10^{12}m^3)=1.49\cdot 10^{19} kg

Also, the asteroid is approximately spherical, so its volume is given by

V=\frac{4}{3}\pi R^3

where R is the radius. Solving the formula for R, we find its radius:

R=\sqrt[3]{\frac{3V}{4\pi}}=\sqrt[3]{\frac{3(3.32\cdot 10^{12}m^3)}{4\pi}}=9256 m

So now we can use eq.(1) to find the escape velocity:

v=\sqrt{\frac{2(6.67\cdot 10^{-11})(1.49\cdot 10^{19}kg)}{9256 m}}=463.4 m/s

3 0
3 years ago
Are we really real or are we all an allusion?Do you believe in past lives?How do you think you died in past life?
OLEGan [10]

Answer:

I don't even know

Explanation:

Shower thoughts

8 0
2 years ago
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Explain how the technology of doing math and adding numbers has changed over time this is science
gizmo_the_mogwai [7]
It has changed because earlier in the days people couldn't even figure out if the world was flat or round but now we can add decimals and multiply fractions.
7 0
3 years ago
To practice Problem-Solving Strategy 27.1: Magnetic Forces. A particle with mass 1.81×10−3 kgkg and a charge of 1.22×10−8 CC has
Roman55 [17]

Answer:

The magnitude of the acceleration  is a = 0.33 m/s^2

The direction is - \r k i.e the negative direction of the z-axis

Explanation:

 From  the question we are that

       The mass of the particle m = 1.8*10^{-3} kg

         The charge on the particle is q = 1.22*10^{-8}C

         The velocity is \= v = (3.0*10^4 m/s ) j

        The the magnetic field is  \= B = (1.63T)\r i + (0.980T) \r j

The charge experienced  a force which is mathematically represented as

         

                    F = q (\= v * \= B)

    Substituting value

         F = 1.22*10^{-8} (( 3*10^4 ) \r j \ \ X \ \  ( 1.63 \r i  + 0.980 \r j )T)

            = 1.22 *10^{-8} ((3*10^4 * 1.63)(\r j \ \  X \ \  \r i) + (3*10^4 * 0.980) (\r j \ \ X \ \  \r  j))

            = (-5.966*10^4 N) \r k

Note :

           i \ \ X \ \ j = k \\\\j \ \  X  \ \ k = i\\\\k  \ \ X \ \ i = j\\\\j \ \ X \ \ i = -k \\\\k \ \  X  \ \ j = -i\\\\i  \ \ X \ \ k = - j\\

Now force is also mathematically represented as

        F = ma

Making a the subject

      a = \frac{F}{m}

   Substituting values

     a =\frac{(-5.966*10^4) \r k}{1.81*10^{-3}}

        = (-0.33m/s^2)\r k

        = 0.33m/s^2 * (- \r k)

6 0
3 years ago
Will Give Brainliest and 25 Points
Vikentia [17]

This is the Doppler effect.

1. As the sound leaves the horn the sound waves are at first close to each other and as they move outwards they become further apart. The closer the sound waves are the louder the noise.

As the car gets the closer the sound waves get closer, so the horn becomes louder.

2. As the horn moves away, the sound waves become less frequent, causing the pitch to get lower.

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